The concepts described herein should give you an idea how you might improve the genetics of your breeding stock over time. Implementing a breed development program using selection based on desirable traits is more difficult than being indifferent to what nature does naturally. While the information may be new to you, you should not rely solely on the oversimplified information given. Breed development is based on many factors, some you cannot control, but if we are to undertake the serious actions of breeding it should only be done by those wishing to make improvements rather than litters.
When it comes to understanding genetics in relation to dog breeding, there are two basic terms to understand, genotype and phenotype.
Genotype describes the complete set of genes inherited by an individual. The difficult concept to understand is that while some dogs carry a genotype to express visually distinct characteristics, they do not come to fruition.
Phenotype are traits expressed visually; what the dog looks like, height, body structure, muzzle color, top-line and the like.
Traits are expressed or hidden based on the presence or complete absence of a dominant or recessive allele.
Alleles are the particular codes in your dog’s genotype that become expressed or hidden through their phenotype. When it is expressed in their phenotype it is said to be "dominant" and your dog possess at least one dominant allele such as D,_. However, without knowing what the other allele is, it could be a recessive allele in which case it said to be "heterozygous" because it does not have a matching pair. An example would be: D,d. However, you may have a "homozygous" "recessive" phenotype expressed because it has a matching pair and it is absent of any dominant allele, in which case it looks like: d,d.
Your dog’s 39 paired chromosomes contain their inherited biological information and what your dog will ultimately look like. These chromosomes consist of tightly wound DNA molecules and have functioning segments of genes attached along the string of DNA called "alleles". There is a multitude of other important terms to understand, but quite frankly, if you made it this far this is an area of interest to you and you should research more in books, scientific journals or college courses.
This picture is not helpful in its own right, but considering these paired 38 autosomes plus a sex linked allosome X and Y, account for a detailed listing of over 1,000 canine diseases, it is easier to understand the multitude of things that can go wrong before your pet is even born!
A detailed listing of over 1,000 canine diseases, and descriptions of each, appearing in the database of inherited diseases in dogs can be found at,
The canine genome has approximately 20,000 genes across 39 chromosomes ( 38 autosomes plus sex linked allosome, X and Y ) and account for a particular dog breeds DNA and ultimately its phenotype. Surprisingly, 14,200 genes originate from a single gene of a last common ancestor between dog, human, and mice ( Hitte 2005 )! We share 71% of our genetic makeup with mice and dogs! Within each breed of dog, a range of 399 to 805 alleles can be found, with an overall average of 605 alleles.
Today there are more than 400 genetically distinct breeds of domestic dogs identified into 4 basic groups; Ancient Asian, Herding,Modern Hunting, and Mastiff. Yet relatively little progress has been made on systematically identifying which regions of the canine genome have been influenced by selective breeding during the natural history of the dog. It is said that nucleotide diversity among all dog breeds accounts for 27.5% of genetic variance. Conversely, the degree of genetic diversity within a particular breed is only 5.4% ( Parker 2004 ). So the difference in genetic makeup between a Chihuahua and a Boerboel is 27.5%, the difference between your Boerboel and my Boerboel's genetic makeup is 5.4%.
Recent mitochondrial DNA ( mtDNA ) sequence analysis has brought light to the domestication of dogs as well as the number of founding groups. MtDNA analysis offers a unique perspective on evolutionary history because the mitochondrial genome is maternally, or female, inherited. A surprise finding of the analysis implies a domestic dog origin of 40 to 135 thousand years ago ( Vila et al. 1997; Savolainen et al. 2002 ), exceeding the previous 15,000yr old archaeological record of dogs. These early dogs may not have been recognized as domestic before 15,000yrs ago because of their physical similarity to gray wolves. Instead, the phenotype of domestic dogs beginning about 15,000yrs ago may indicate a change in selection pressures associated with the transition of humans from hunter to a more sedentary lifestyle of gatherer ( Wayne 2006 ).
Quantitative Genetics looks mathematically at how complex traits are inherited. That is, many genetic and environmental factors involved in determining the value of a trait. If you think about it, most of the traits we are really interested in for dogs are quantitative traits. One of the important results we get from this kind of number crunching, and one that you will see a lot in genetics literature, is heritability. Heritability is the amount of variance in the trait that is due to genetic factors. Variance is another way to describe the difference between individuals throughout their phenotype. A simple equation can be used to describe the result:
P = G + EF ( Phenotype = Genotype + Environmental Factors ).
Inheritance is the reason people buy a "pure bred" dog. They are looking for consistent and known traits, characteristics, and appearances of what a responsible breeding program brings into a cluster of genetic potential. When dealing with a breed development program, and anyone taking the responsibility of breeding should be doing so ONLY to make improvements in a breed, they need to have a plan on HOW and WHERE improvements are to be made. There is a reason HALF of all litters are murdered in animal shelters every year regardless of breed rarity or expense, because back-yard "breeders" and those not in the "know" have no breed development plan and no thought process for animal welfare!
So how do we develop a breed? We first need to understand which traits we want to keep, which could be improved, and which should have no place in a breeding population. Those all come down to inheritance. Once again we need to identify those traits which are dominant and those that are recessive. If we have a trait that we do not want to continue, but it appears the most frequently, ie: dominant, this becomes the easiest to eliminate because we choose individuals that do not show signs of that trait. It is much more difficult to remove a recessive allele from a population. It will only become expressed when paired with another recessive allele as it is "hidden." It is often hidden throughout several generations, at which point it has already been inherited and spread among the breeding population.
The Hardy-Weinberg equation describes how alleles behave in a population from generation to generation. It is an oversimplification in that it assumes that mating is random, mutation does not occur, environmental factors are not in play, and there is no selection, natural or otherwise. Not realistic in a responsible breed development program.
In a system with two alleles, where the expression of p is dominant and the expression of q is recessive, the equation states:
p2 + 2pq + q2 = 1
p2 is the expression of p homozygous dominant.
2pq is the expression of pq heterozygous where p is dominant, but has a recessive q. There is a 2 in front because you can get this genotype in two ways:
a) p from the mother and a q from the father, or
b) p from the father and a q from the mother.
q2 is the frequency of q homozygous recessive. Now it becomes expressed in the phenotype because there is no dominant p to hide it in the genotype.
Another way to show this is with a Punnett Square (this will be described in detail later).
P PP Pq
q Pq qq
So, by adding across the columns because alleles are paired, we get: PP, Pq, Pq, qq. You also see a 1:2:1 ratio.
Which is: p2 + 2pq + q2 = 1
If p and q are equal expressed at 0.5 or 50% because you have a 50-50 ratio ( 1:1 ), then the equation looks like:
0.25 + 0.5 + 0.25 = 1
If p is expressed at 0.75 or 75% and q is expressed at 0.25 or 25%, you have a 75-25 ratio ( 3:1 ), then the equation looks like:
0.56 + 0.38 + 0.06 = 1
Now, the practical application of it. Suppose you start with a recessive trait or an undesirable allele that is expressed at 0.25 or 25% of the time, and you decide you will only breed individuals who show no phenotype for that allele over time...
Generation p2 2pq q2
G1 0.25 0.5 0.25
Only individuals from 2 of the 3 possible groups are going to be bred in Generation 2 because the p2 and 2pq columns have no phenotype for the recessive allele. You have 0.33 or 33% that express the recessive trait and .66 or 66% that show the dominant trait, even though 50% of their genotype still maintains the recessive allele and only 25% are pure dominant. The allele frequency for p is now the square root of 0.33, or 0.57. Since p + q = 1, q is now 0.43, and
G2 0.33 0.49 0.18
When continuing this way to eliminate all the q2's we know will express the trait we want to eliminate, we get… Notice how the further we go, the less the last column decreases each time.
G3 0.40 0.47 0.13
G4 0.45 0.44 0.11
G5 0.51 0.41 0.08
G6 0.55 0.38 0.07
G7 0.60 0.35 0.05
G8 0.63 0.33 0.04
G9 0.68 0.29 0.03
G10 0.70 0.27 0.03
G11 0.72 0.26 0.02
G12 0.73 0.26 0.02
A better visual demonstration involves a lot more work, but is an interesting way to describe inheritance and how the occasional “black sheep” can surface.
Take for example the dominant and recessive gene expression. As you recall, all you need is a dominant gene ( D? ) to have a phenotype expressed. It is of no consequence right now what the other allele is because we have one dominant D that becomes expressed. If we take this specific dog and breed it to a dog that shows no phenotype for trait D, such as homozygous recessive ( dd ), we get:
d D,d d,?
d D,d d.?
So before we do the breeding we have a 50-50 ( 1:1 ) probability that the trait will be passed on because we know for sure we only have one dominant allele ( D? ). We also have a 50% probability that the recessive allele will become expressed ( dd )! Scary!!
Our litter shows that all the puppies show the dominant trait D, which is what we were hoping for! Now we can assume that the unknown allele was the dominant allele D ( DD ):
d Dd Dd
d Dd Dd
The result is 100% Dd for the puppies. Sounds good, but our next breeding if we bred any of the puppies to a dog that is homozygous recessive for the trait ( dd ), it will result in only HALF showing the desired trait we worked to get:
d Dd dd
d Dd dd
What if instead we looked to a breeding partner with similar traits? We can guess that the other dog has at least a heterozygous dominant allele ( D? ) and our know dog has ( Dd ):
D DD Dd
? D? d?
Now we have a 75-25 ( 3:1 ) probability that our dominant trait D will be expressed. Much better than 50%! Why is it that “breeders” use better quality dogs to “clean-up” less quality dogs? By doing so, they at best have a 50-50 probability of a desired outcome. If I said you have a 50-50 chance of making it out of the driveway alive would you go? What if I said you have a 50-50 chance of getting a quality pup from someone careless about their breed development program who charges a lot more money than the local animal shelter for the same thing? Four legs, barks when necessary, and provides unconditional love. Is not that what all dogs provide? You are paying for inherited traits...and by most breeders you will only have a 50% chance of getting a desired trait and that is just one of the averaged 605 alleles that make up your pets phenotype! Now, what if that recessive allele pertains to poor health or over aggression? The expense of that puppy suddenly will cost you much more in vet bills and potential lawsuits!
Now we get down to the important stuff! We know that there are an average of 605 alleles per dog breed, that means a possible 605 different phenotypic combinations. All alleles do not account for outward traits however. Below we will look at just 5.
B,?=Big Head. D,?=Dense Bone. F,?=Flat Top. W,?=Wide Chest. M,?=Muscular Thighs.
The above capital letter indicates the dominant allele. It would take a homozygous recessive allele of each to show a dog of average quality: ( b,b ), ( d,d ), ( f,f ), ( w,w ), ( m,m ). We decide to take a good looking male and breed him to a lesser female in the hopes of "cleaning them up". Much of that mentality is someone that spent a lot of money on a dog and does not want to lose out on their "investment":
B ? / D ? / F ? / W ? / M ?
b Bb b?
b Bb b?
d Dd d?
d Dd d?
f Ff f?
f Ff f?
w Ww w?
w Ww w?
m Mm m?
m Mm m?
The top rows phenotype is: Big Head, Dense Bone, Flat Top, Wide Chest, Muscular Thighs. The side columns phenotype is average for everything.
Before we do the breeding, we have a 50-50 probability of making an improvement through heterozygous dominant alleles ( Dd ) or making a terrible contribution of 50% being homozygous recessive ( dd ). That is, if their offspring breed they WILL pass down a less than desirable trait! Let’s say that we lucked out or did some research and found that the male is homozygous dominant ( DD ), in which case the unknown allele is dominant. We just made an improvement in that we hid all the recessive alleles and thus all the puppies phenotypes are the same ( Dd ). When breeding these offspring we should only look for other animals of similar traits ( Like x Like ) to breed with to make the dominant allele once again homozygous ( DD ). In effect the breeder took one step back to go one step ahead.
Now let’s look at a breeding of Like x Like. That is a breeding pair of similar traits:
B ? / D ? / F ? / W ? / M ?
B BB B?
? B? ??
D DD D?
? D? ??
F FF F?
? F? ??
W WW W?
? W? ??
M MM M?
? M? ??
The top row and side columns phenotype are: Big Head, Dense Bone, Flat Top, Wide Chest, Muscular Thigh.
As you can see, before we even see the results we have a 75% probability that the dominant ( D? ) allele will be expressed and a 25% probability the homozygous recessive ( ?? ) allele will be expressed, in which case the breeder should never use for a breeding. We also have a 25% probability of a homozygous dominant ( DD ) allele to continue a successful breed development program! A Like x Like breeding should be the preferred method because statistically there is a 75% probability of improvement.
Now let’s look at another method of using good breeding stock to strengthen other breeding stocks weak areas:
B ? / D ? / f f / W ? / m m
b Bb b?
b Bb b?
d Dd d?
d Dd d?
F Ff Ff
? f? f?
w Ww w?
w Ww w?
M Mm Mm
? m? m?
The top rows phenotype is: Big Head, Dense Bone, sloped top, Wide Chest, weak thighs. The side columns phenotype is: small head, small bone, Flat Top, weak chest, Muscular Thighs.
Here we still end up with a 50-50 probability of improvement. At best, if our unknown allele was dominant, we took a homozygous dominant phenotype ( DD ) and turned it into a heterozygous dominant ( Dd ). While the puppies would come out all expressing the desired trait if this were true, we took a step back once again to take a step forward, and that is only if we choose a "Like x Like" breeding with a desired puppy.
PUNNETTE SQUARE and CHI SQUARE test.
In genetics we often attain results that are close to expected results, but not identical. If before we do a breeding between a tan color Boerboel and a brindle color Boerboel we think that the tan color will be dominant, we have learned that we have a 50-50 probability the puppies will be tan with a heterozygous dominant allele ( Dd ):
d Dd dd
d Dd dd
We can use a statistical test called the Pearson's Chi-Test ( pronounced "khi" like "pie") to check the observed to expected ratios. If P ( probability ) from the calculated value is greater than 5%, we accept the observed results as true and not left to chance. In order to find the number we use the equation:
Where E is the expected trait number, O is the observed trait number, and Σ means the "sum of all". Using Mendelian ratios to test a heterozygote hypothesis ( Dd ) that our tan breeding member possess a recessive allele, we will let a tan color trait be expressed in ( Dd ) and a brindle color trait be expressed with homozygous recessive ( dd ). In a repeat breeding, our total puppy count is 32. We should have a 50-50 probability of ( Dd ):( dd ). Instead we have 18 tan and 14 brindle offspring. Was our hypothesis wrong?
Trait O E (O-E)² (O-E)²/E
Tan 18 16 4 0.25
Brindle 14 16 4 0.25
X equals 0.25+0.25= .5
Now we use a Critical Values of X Distribution Table which will give us the P ( probability ) value we want. It uses degrees of freedom or df to indicate the number of independent variables in the data minus 1. Our df number is 2 ( tan and brindle ) -1 = 1. When looking at the table below, we find our X = .5 on the df line between columns marked 0.5 and 0.1.
0.9 0.5 0.1 0.05 0.025 P
1 0.016 0.455 2.706 3.841 5.024
2 0.211 1.386 4.605 5.991 7.378
Since our P is greater than 0.05 or 5% our hypothesis is true! Our Chi Test only works with actual observable numbers not percentages and has to have more than ten in the outcome to be valid.
Calculating the probabilities or expected frequencies of specific phenotypes/genotypes from a particular breeding involves the application of simplified statistical rules. We will look at the product rule and the sum rule.
The Product Rule states that the probability of independent events occurring together is the product of the probabilities of the individual events. It focuses on outcomes A "and" B. An over simplified example would be the probability of rolling a pair of 4's on dice. We have a 1 in 6 chance per die so: P( of two 4"s ) = 1/6 x 1/6 = 1/36.
If we use the Punnett Square above with the 5 traits of: Big Head ( Bb ), Dense Bone ( Dd ), Flat Top ( Ff ), Wide Chest ( Ww ), Muscular Thigh ( Mm ) and do a breeding of "Like x Like", each trait has a 3/4 probability of expression and a 1/4 probability of expressing the recessive trait. We want to know what the probability is of having a homozygous recessive ( dd ) puppy for all traits in this particular breeding. To do so we must first calculate the probability for each outcome, and then multiply them all:
average head(1/4), average bone(1/4), slope top(1/4), weak chest(1/4), weak thigh (1/4)
P = (1/4) x (1/4) x (1/4) x (1/4) x (1/4) = 1/1024.
The Sum Rule states that the probability of either of two mutually exclusive events occurring is the sum of their individual probabilities. The focus is on the outcome A or B. An oversimplification of the rule involving dice would be the probability of rolling a 4 or a 5 on dice. P( of a 4 or 5 ) = 1/36 + 1/36 = 2/36 = 1/18.
As stated before, using the Punnett Square and the 5 traits, our first example of taking a good looking male and breeding him to a lesser female in the hopes of "cleaning them up", we want to know the probability of having similar traits of a Big Head and Wide Chest or Dense Bone and Muscular Thighs.
Big Head (1/2), Wide Chest (1/2), Dense Bone (1/2), Muscular Thighs (1/2
P= [ ( 1/2 x 1/2 ) + ( 1/2 x 1/2 ) ] = ( 1/4 )+( 1/4 ) = 2/4 = 1/2
You see that our previous probability of of the drawn out Punnett Square is true AND we were able to incorporate both probability rules!
EFFECTIVE LINE BREEDING.
There is a favorite breeding system used by successful breeders of a variety of animals. It can produce superior quality animals of desired traits IF the male selected is an outstanding example, nearly faultless, and has proven himself to be able to pass on his phenotype. It states "Let the sire of the father be the grandsire of the mother on the mother's side." I will show a pedigree diagram to illustrate.
Another key when looking at a pedigree is to equate how much of a role the puppy’s relatives play in their potential phenotype. Too often I see people touting their lineage from grandfather, great-grandfather, or higher just because they "scored" high on a particular appraisal. Since then their offspring are of little note. Truth be told, the greatest contributors are the parents, and they only contribute a potential 25% each! Even still, without a scientific approach and attention to detail, breeders are only capable of attaining a 50% expected result. In blind faith, a client pay's good money based on what a breeder tells them or shows them, most likely in an appraisal score, in the hopes of having a healthy, structurally sound dog.
When looking at genetic inheritance, we can see that within 5 generations we are able to completely change what was, to what is. Science has shown us that we can fix hips, size, color, aggression and appearance within these 5 generations with selection for desired traits. The Silver Fox experiment, the bulldog, inter-race relationships, breed hybrids, these are but a few examples where effective change in phenotype can be seen.
Generation. Gene Contribution This Generation. Each Parents Genetic Contribution.
F4 This Dog 100 %
F3 Parents 50 % 25 % X 2
F2 Grand Parent 25 % 6.25 % X 4
F1 Great GrandP 12.5 % 1.56 % X 8
P GreatGreatGP 6.25 % 0.39 % X 16
The real difficulty in choosing which dog's to breed, especially if they are from a different kennel and different bloodline is how to assess their genetic potential. Inbreeding and line breeding ( one-in-the-same ) provide the genetic material to better build a pedigree, BUT there are terrible consequences involved for everyone if the breeder does not know how to master and manipulate the physical traits of the mating pair. It demands extreme discrimination, something the vast majority of breeder’s lack. People associate inbreeding with genetic defects. While it is true that defects caused by recessive alleles do surface in inbred populations, inbreeding does not create harmful recessive alleles; they must already have been present in a population. It therefore increases the chance of damaging alleles becoming prevalent and expressing themselves sooner than in random breeding’s.
For simplicity sake, we will look at the heritability of the sex linked allosome X and Y to prove the favored breeding system of those that undertake the task. The X chromosome is present in both males ( X,Y ) and females ( X,X ) and is comprised of a very interesting pattern of ancestor contribution. Males are given their only X chromosome from their mother, but women receive one from each parent. Females can pass along an X chromosome which contains information from both their parents, but males can only pass along an X that is a virtual copy of the one they received from their mothers.
F3 F2 F1 P
X,X1* / X1*,Y* and
X,X2 / X2,Y* X1*,X2
X1*,X3 / X3,Y*
X1*,X4 / X4,Y* and X3,Y1
X3,X4 / X4,Y1 and
X3,X5 / X5,Y1 X4,X5
X3,X7 / X7,Y*
X1*,X1* / X1*,Y* and
X6,X7 / X7,Y2 and
X6,X8 / X8,Y2 X7,X8
X7,X1* / X1*,Y2
X7,X9 / X9,Y2 and
X1*,X9 / X9,Y* X1*,Y*
X1*,X10 / X10,Y* and
You can see and follow the X1*,Y* allosome through three generations and how we end up with one male pup identical to the "sire of the father be the grandsire of the mother on the mother's side" breeding system. We also get a super female X1*,X1* and another male with X7,Y*. That makes 75% of the offspring in possession of inherited allosomes among the three generations. This is why a breeding system of this nature can be VERY successful; however, if a less than perfect male is used we can also see that no improvements are made and faults are exemplified! Those highlighted in blue were the chosen parents per generation.
You can also set up the same scenario with half-brother x half-sister mating’s on the sire side or mother's side. We can also do a father x daughter combination and get an identical male to the father, an identical female to the daughter, a super female ( X1,X1 ), and a male with the same Y as the father. This should not be done whimsically! Instead it should only be attempted at the end of a well-planned pedigree.
COPYRIGHT ADRIAN BOERBOELS 2007-2017. ALL RIGHTS RESERVED.